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sodium metal reacts with water to produce hydrogen gas according to the followin

ID: 704697 • Letter: S

Question

sodium metal reacts with water to produce hydrogen gas according to the following equatuion: 2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2 (g). in an experiment. the hydrogen gas is collected over water in a vessel where the total pressure is 745 torr and the temperature is 20 c, at which temperature the vapor pressure of water is 17.5 torr. under these conditions, what it the partial pressure of hydrogen? if the wet hydrogen gas formed occupied a volume of 8.04 L, what number of moles of hydrogen is formed?

Explanation / Answer

2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2 (g).

the total pressure is 745 torr

the vapor pressure of water is 17.5 torr

the partial pressure of hydrogen   = the total pressure - the vapor pressure of water

                                                     = 745-17.5   = 727.5 torr

PH2   = 727.5 torr   = 727.5/760   = 0.96atm

V    = 8.04 L

T = 20+273 = 293K

PV = nRT

n    = PV/RT

       = 0.96*8.04/0.0821*293   = 0.32 moles

The number of moles of hydrogen is 0.32 moles >>>>answer